   Chapter 18, Problem 142CP

Chapter
Section
Textbook Problem

# The overall reaction in the lead storage battery is Pb ( s )   +   PbO 2 ( s )   +   2 H + ( a q )   +   2 HSO 4 − ( a q )   →   2 PbSO 4 ( s ) +   2 H 2 O ( l ) a. For the cell reaction ∆H° = −315.9 kJ and ∆S° = 263.5 J/K. Calculate at −20.°C. Assume ∆H° and ∆S° do not depend on temperature.b. Calculate at – 20.°C when [HSO4−] = [H+] = 4.5 M.c. Consider your answer to Exercise 71. Why does it seem that batteries fail more often on cold days than on warm days?

(a)

Interpretation Introduction

Interpretation:

The overall reaction taking place in lead storage battery is given. various questions based on the given battery have to be answered.

Concept introduction:

The device that contains one or more electrochemical cells that is used to provide electrical energy to the different electrical devices is called a battery.

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature, the above equation specifies as,

E=E°(0.0591n)log(Q)

The relationship between Gibbs free energy change and cell potential is given by the formula,

ΔG°=nFE°cell

The relation between ΔG° , ΔH° and ΔS° is given as,

ΔG°=ΔH°TΔS°

To determine: The value of E° at the given reaction conditions.

The value of E° is 1.98V_ .

Explanation

Given

The value of ΔH° is 315.9kJ .

The value of ΔS° is 263.5J/K .

The given temperature is 20°C .

The conversion of kJ to J is done as,

1kJ=1000J

Conversion of 315.9kJ to J is done as,

315.9kJ=315.9×1000J=-315.9×103J

The conversion of degree Celsius (°C) into Kelvin (K) is done as,

T(K)=T(°C)+273.15

Hence, the conversion of 20°C to K is done as,

T(K)=20°C+273.15=253.15K

The reaction taking place at the cathode is,

PbO2+HSO4+3H++2ePbSO4+2H2O

The reaction taking place at the anode is,

Pb+HSO4PbSO4+H++2e

The reaction taking place in lead-storage battery is,

Pb(s)+PbO2(s)+2H+(aq)+2HSO4(aq)2PbSO4(s)+2H2O(l)

The relation between ΔG° , ΔH° and ΔS° is given as,

ΔG°=ΔH°TΔS° (1)

Where,

• ΔG° is the standard Gibbs free energy change

(b)

Interpretation Introduction

Interpretation:

The overall reaction taking place in lead storage battery is given. various questions based on the given battery have to be answered.

Concept introduction:

The device that contains one or more electrochemical cells that is used to provide electrical energy to the different electrical devices is called a battery.

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature, the above equation specifies as,

E=E°(0.0591n)log(Q)

The relationship between Gibbs free energy change and cell potential is given by the formula,

ΔG°=nFE°cell

The relation between ΔG° , ΔH° and ΔS° is given as,

ΔG°=ΔH°TΔS°

To determine: The value of E at the given concentration of ions.

The value of E is 2.05V_ .

(c)

Interpretation Introduction

Interpretation:

The overall reaction taking place in lead storage battery is given. various questions based on the given battery have to be answered.

Concept introduction:

The device that contains one or more electrochemical cells that is used to provide electrical energy to the different electrical devices is called a battery.

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature, the above equation specifies as,

E=E°(0.0591n)log(Q)

To determine: The reason of failure of batteries in cold days than on warm days.

The decrease in value of cell potential with decrease in temperature leads to the failure of batteries in cold days than on warm days.

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